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RC in Parallel

Discussion in 'Electrical Engineering' started by Peter, Feb 17, 2005.

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  1. Peter

    Peter Guest

    I'm trying to figure something out and so far no one has come up with an

    If I have a voltage divider network (Just 2 resistors) and I bypass the
    bottom one with a cap with an AC source in. How do I figure out the

    My goal is the following: put a sine wave in with say 4v p-p in @1kHz, have
    an R1 and R2 (equal value) with a bypass cap on R2 (the other side of R2 is
    connected to ground) and say I get 1v p-p out (the middle of R1 and R2 is my
    output). I want to take that circuit and replace it with just R1 and R2
    (keeping R1 the same value) and get the same amplitude out. I don't care
    about phase right now.

    Anyone have any ideas?
  2. Roy Q.T.

    Roy Q.T. Guest

    I've had no need to figure anything remotely close to this in years but;
    Don't you have to calculate XC or capacity reactance to get Z impedance
    in an RC network ?

    good luck !

    Roy ~ E.E.Technician
  3. Guest

    If you already know what you want for Vout (1 volt)
    and you know that Vin is 4 volts, why do you care
    what the cap is doing, since it won't be in the
    new circuit? You know its presence in the existing
    circuit with equal Rs causes the the thing to
    divide by 4.

    Anyway: Vout = Vin*NewR2/(R1+NewR2) where Vout appears
    across the NewR2

    If you want to know the impedance of your existing
    R2 and cap, solve the above equation for the new
    R2. Then solve for Zc with R2new = R1*Zc/(R1+Zc).
    Or figure the capacitave reactance at 1 khZ
    Xc = 1/2*pi*f*C. In this case, Xc and Zc are
    the same.

  4. Do you mean the impedance as seen by the supply? In which
    case, simply work out the impedance of one resistor in
    series with a parallel combination of the capacitor, the
    other resistor (and what ever load you might want to work it
    out for).

    Do you mean the impedance as seen by a load? In which case
    it is the impedance of the parallel combination of the
    resistor and capacitor, in series with the resistor (and the
    impedance of ths source, if not considered negligable).
    The voltage source will see R1 in series with the parallel
    combination of R2b and your load impedance. Assuming that
    your load impedance is high enough to be insignificant, this
    reduces to R1 in series with R2b. If you can't work out the
    value of R2b - bearing in mind that the same current flows
    through R1 and R2b but that the voltage across R1 is three
    times the voltage across R2b, then you want to take up
    needlework and/or cookery and leave electrickery to us girls.
  5. Bob Eldred

    Bob Eldred Guest

    Why do you have the resistor bypassed if you are trying to make a voltage
    divider? In a high frequency attenuator, both resistors are in parallel with
    capacitors and have their time constants equal like in a scope. But, I don't
    understand what you are trying to do.
  6. meqshesher

    meqshesher Guest

    presented to the previous stage or the next stage?

    "Bypass" means AC short, so the impeadance presented to the next stage will
    be low (looking mostly at the cap impeadance)

    You have 1/2 voltage out R2/(R1+R2) (no cap)
    Now your divider is 1/4 so Z of the cap is about 1/3 of R (estimate)
    So take the cap out? and have a voltage division of 1/4?

    1/4 = R2/(R1+R2) so 4*R2=R1+R2 or R1= 3*R2

    So if R1 is 1k then R2= 330 ohms and presented Z to previous stage is
    about 1.3k and Z presented to next stage is about 250 or so. (need Z's of
    the stages to have real answers)
  7. Peter

    Peter Guest

    The reason I'm trying this circuit is because I was trying to understand
    bypass caps in a common emitter transistor circuit. Everytime I tried
    figuring out the impedance looking into the base, I didn't get what I

    So I built this circuit to learn from it then use what I learn and apply
    that back into a common emitter circuit.
  8. Guest

    That's a different kettle of fish. Try this site to see if it helps:

  9. Jimmie

    Jimmie Guest

    How far off are you calculation? Remember to include the base /emitter
    junction resistance. Mostly this value is insignificant unless the the total
    emitter resistance is by passed. If the emitter resistor is completely
    bipassed this may have the most significant effect on the input impedance
    depending on the circuit design.
  10. Peter

    Peter Guest

    Well that's why I'm using THIS circuit so I don't have to worry about all
    the factors of the transistor.

    If I can figure out how to go from a 2 resistor series circuit with the
    bottom resistor bypassed by a cap and the middle of the two resistors as my
    Vout to just a 2 resistor series circuit with ONLY changing the bottom value
    resistor to the equivilent of the R and C combiation and get the SAME Vout,
    then I will know for sure that my emitter is all set and any variations I
    get in calculations will be due to temp or whatever.

    This is why this circuit is important. I don't care about phase. I ONLY care
    about how to go from 1kHz 4Vp-p with both resistors equal to 1k and the
    bottom resistor bypassed with a 1uf cap while I get 610mVp-p and then take
    the R and C to get a total impedance, remove the cap and install my new
    calculated value reistor so I can get the 610mVp-p output with JUST two

    when I calculate the bottom resistor, I get 157ohms, but that does not equal
    610mVp-p on the output. The value that does give me the same amplitude is
    178ohms, but I have NO idea how to calculate that value.

    Thanks again
  11. Peter

    Peter Guest


  12. What's the big problem?

    Vr1=(4-0.610) voltage across r1
    r1=1k a given
    V/R=I (0.00339) find current through r1
    Vr2=(0.610) a given
    V/I=R find the resistance that will drop 0.61V at 3.39mA

    you said it works with a 178 ohm
  13. Peter

    Peter Guest

    OK. figure out the resistance of the R and C without using any voltages
  14. Guest

    This has been answered before. In case you missed it:
    Xc = 1/2*pi*F*C
    That gives you the capacitve reactance in ohms.
    If you already know the total resistance of R and C in parallel:
    Rtot = (R*Xc)/(R + Xc)
    Solve for R

    If you know R but do not know Rtot, the equation is the same:
    Rtot = (R*Xc)/(R + Xc)
    Solve for Rtot

  15. daestrom

    daestrom Guest

    Sorry, this doesn't give the correct results. Xc is displaced 90 degrees
    from R. A 10 ohm resistor and a capacitive reactance of 10 ohms in parallel
    with it does *not* give you (10*10)/(10+10) = 5 ohms impedance.

    With parallel components, to combine them it is useful to calculate the
    individual conductance and susceptance, then add them (being sure to account
    for resistance vs. reactance), then convert the resulting admittance back to
    impedance (you just got to love all these terms ;-)

    For example, a 10 ohm resistor has a conductance of 0.1 mho. Similarly, the
    capacitor with 10 ohm reactance has a susceptance of 0.1 mho. These combine
    to give the complex admittance

    For parallel circuits....
    admittance = sqrt (conductance^2 + susceptance^2)
    = sqrt (0.1^2 + 0.1^2) = 0.1414 mho

    So the impedance is 1/0.1414 = 7.07 ohm

    Lumped all together, for a resistance R and capacitive reactance Xc....

    Z = 1 / (sqrt( (1/R)^2 + (1/Xc)^2)),,sid44_gci212333,00.html

    If the resistance or reactance is significantly larger than the other, your
    formula will give you *approximately* correct results. But if the ohmic
    values are near each other, the proper calculation is what I just described.

  16. Guest

    Thanks - you are correct and my answer was wrong.
    Started thinking DC (obviously) when its AC. I hate
    it when I do that.
  17. Roy Q.T.

    Roy Q.T. Guest

    };-) Okayyy ?
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